Disintegration and Measure-zero sets Suppose that $X,Y$ are Radon spaces, $\mu$ is a Borel probability measure on $X$, and $\nu$ is a $\sigma$-finite Borel measure on $Y$.  Fix some non-empty $A\subseteq X\times Y$ such that
$$
\mu\otimes \nu (A)=0.
$$
Can we then conclude (somehow by disintegration) that the projection of $A$ onto $Y$ are of $\nu$-measure $0$?
 A: Well, if $A=N\times B,$ where $N$ is a $\mu$-null set and $B$ is any Borel subset of $Y$, then $\mu\otimes \nu(A)=0,$ but the measure of $B$ can be whatever we want.
A: Assuming that $X$ is Hausdorff, then this statement is true for all $A$ if and only if one of the following holds:


*

*$X$ is countable, and the measure $\mu$ assigns positive measure to every point; or

*$\nu$ is the zero measure.
For one direction: suppose $X$ is countable, so $X = \{x_1, x_2, \dots\}$, and every point of $X$ has positive measure.  Then we can write $A = \bigcup_n \{x_n\} \times A_{x_n}$, where $A_{x} = \{y : (x,y) \in A\}$ is the section of $A$ through $x$.  By Fubini's theorem, $A_{x}$ is measurable for almost every $x$, which in this case means every $x$, and we have $0 = (\mu \otimes \nu)(A) = \sum_n \mu(\{x_n\}) \nu(A_{x_n})$.  This implies $\nu(A_{x_n}) =0$ for all $n$.  Now the projection $\pi(A)$ is given by $\bigcup_n A_{x_n}$ which is therefore measurable and has measure zero.
Of course if $\nu = 0$ then the result is trivially true.
For the converse: if $Y$ has a set of positive measure and $X$ has a nonempty set of zero measure, then as WoolierThanThou showed, the product of those sets  gives an $A$ for which the claim does not hold.  So either $Y$ has no sets of positive measure, which is to say $\nu =0$, or else the only measure-zero set of $X$ is empty.  Suppose the latter happens.  Since $X$ is Hausdorff, singletons are closed and thus measurable, so every singleton must have positive measure.  Then in order for $\mu$ to be finite or even $\sigma$-finite, there can only be countably many singletons, which is to say $X$ is countable.
